tried this set(i) <= set(cords) but got error TypeError: unhashable type: 'list' I can loop through the list but i guess there is a better way
#🔒 how do i check if all elements of a list are there in another list
misty cargo
gusty adderBOT
tried this `set(i) <= set(cords)` but got error `TypeError: unhashable type: 'li...
@misty cargo
Python help channel opened
shut shale
tried this `set(i) <= set(cords)` but got error `TypeError: unhashable type: 'li...
what are the contents of your list (which you try to convert to set ) ?
misty cargo
a list of 2 elemnts [x:int,y:int]
fast dock
If you have something like this:
a = [1,2,3]
b = [1,2,3,4,5]
you can check it with:
a in b
misty cargo
If you have something like this:
a = [1,2,3]
b = [1,2,3,4,5]
you can check it ...
really thanks
shut shale
!e
a = [1,2,3]
b = [1,2,3,4,5]
print(a in b)
:white_check_mark: Your 3.12 eval job has completed with return code 0.
False
you are trying to check if [1,2,3] as a whole is inside of the other list
fast dock
Oh sorry, my bad
shut shale
!e
a = [1,2,3]
b = [[1,2,3],4,5]
print(a in b)
:white_check_mark: Your 3.12 eval job has completed with return code 0.
True
Yeah right, then you can use:
set(a).issubset(b)
misty cargo
Yeah right, then you can use:
set(a).issubset(b)
doesnt work
shut shale
doesnt work
!e
a = [1,2]
b = [1,2,3]
print(set(a).issubset(b))
:white_check_mark: Your 3.12 eval job has completed with return code 0.
True
but for me
a=[[1,1],[0,0]]
b=[[1,1],[0,0],[1,5]]
ahh , you have list inside of lists , that changes things
topaz wave
Yeah right, then you can use:
set(a).issubset(b)
That's the same as <=.
brisk robin
but for me
```
a=[[1,1],[0,0]]
b=[[1,1],[0,0],[1,5]]
```
integers are hashable, which means they can be in sets. Lists are not hashable, so they can't be in sets.
misty cargo
integers are hashable, which means they can be in sets. Lists are not hashable,...
thats fine but how do i check if a is in b ? looping though a and check if each element is in b is the only option
olive sierra
You can use the all built-in too, to make it clearer
!e
a=[[1,1],[0,0]]
b=[[1,1],[0,0],[1,5]]
print(all(x in b for x in a))
:white_check_mark: Your 3.12 eval job has completed with return code 0.
True
thanks
topaz wave
thats fine but how do i check if `a` is in `b` ? looping though `a` and check if...
if you were to turn those sublists into tuples, sets would work.
misty cargo
that can be done
olive sierra
!e
```py
a=[[1,1],[0,0]]
b=[[1,1],[0,0],[1,5]]
print(all(x in b for x in a))
``...
Do note for this approach though, that [0, 1] != [1,0]. I'm not sure that's what you want?
misty cargo
Do note for this approach though, that [0, 1] != [1,0]. I'm not sure that's what...
ya that is what i want
would your code be more efficient than this
set(map(tuple, i)).issubset(map(tuple, cords))
!e
Oh then maybe the tuple approach isn't for you, py print(set((0, 1)) == set((1,0)))
gusty adderBOT
!e
Oh then maybe the tuple approach isn't for you, ```py
print(set((0, 1)) == se...
:white_check_mark: Your 3.12 eval job has completed with return code 0.
True
!e
print(set([(0, 1)]) == set([(1,0)]))
:white_check_mark: Your 3.12 eval job has completed with return code 0.
False
this is what i want
olive sierra
Oh yeah mb
misty cargo
!e
print(set([(1,0)]) == set([(1,0)]))
:white_check_mark: Your 3.12 eval job has completed with return code 0.
True
Then yours is better (I think?) because tuples are immutable so there isn't as much overhead? Maybe? Idk though
Oh but you're converting to tuples on the fly, so that might be a bit more work if the list is huge
Can't you have them always be tuples, since the length is always 2?
misty cargo
Oh but you're converting to tuples on the fly, so that might be a bit more work ...
hmm idk I usally never use tuples as its not mutable but might use in this context as i wont change the cords
misty cargo
Can't you have them always be tuples, since the length is always 2?
ya they are cordinates of my point
thankyou all
